hdu1532 sap模板

之前在做好多题 都用dicnic和E-K算法 

结果连续T了好几天一道题也没过

直到知道还有SAP算法

就直接摘了kuangbin的模板

在网上找了好多的模板 发现还是kuangbin的最好用

用裸题hdu1532上代码

Drainage Ditches

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18235    Accepted Submission(s): 8592

Problem Description

Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Thus, Farmer John has built a set of drainage ditches so that Bessie's clover patch is never covered in water. Instead, the water is drained to a nearby stream. Being an ace engineer, Farmer John has also installed regulators at the beginning of each ditch, so he can control at what rate water flows into that ditch. 
Farmer John knows not only how many gallons of water each ditch can transport per minute but also the exact layout of the ditches, which feed out of the pond and into each other and stream in a potentially complex network. 
Given all this information, determine the maximum rate at which water can be transported out of the pond and into the stream. For any given ditch, water flows in only one direction, but there might be a way that water can flow in a circle. 

 

Input

The input includes several cases. For each case, the first line contains two space-separated integers, N (0 <= N <= 200) and M (2 <= M <= 200). N is the number of ditches that Farmer John has dug. M is the number of intersections points for those ditches. Intersection 1 is the pond. Intersection point M is the stream. Each of the following N lines contains three integers, Si, Ei, and Ci. Si and Ei (1 <= Si, Ei <= M) designate the intersections between which this ditch flows. Water will flow through this ditch from Si to Ei. Ci (0 <= Ci <= 10,000,000) is the maximum rate at which water will flow through the ditch.

 

Output

For each case, output a single integer, the maximum rate at which water may emptied from the pond. 

 

Sample Input

5 41 2 401 4 202 4 202 3 303 4 10

 

Sample Output

50

#include<algorithm>#include<iostream>#include<cstdlib>#include<cstring>#include<cstdio>#include<string>#include<stack>#include<queue>#include<cmath>#include<stack>#include<list>#include<map>#include<set>typedef long long ll;using namespace std;const int MAXN=20010;//点数的最大值const int MAXM=880010;//边数的最大值const int INF=0x3f3f3f3f;struct Node{    int from,to,next;    int cap;}edge[MAXM];int tol;int head[MAXN];int dep[MAXN];int gap[MAXN];//gap[x]=y :说明残留网络中dep[i]==x的个数为yint n;//n是总的点的个数，包括源点和汇点void init()  //在main函数的开头写init别忘了!{    tol=0;    memset(head,-1,sizeof(head));}void addedge(int u,int v,int w){    edge[tol].from=u;    edge[tol].to=v;    edge[tol].cap=w;    edge[tol].next=head[u];    head[u]=tol++;    edge[tol].from=v;    edge[tol].to=u;    edge[tol].cap=0;    edge[tol].next=head[v];    head[v]=tol++;}void BFS(int start,int end){    memset(dep,-1,sizeof(dep));    memset(gap,0,sizeof(gap));    gap[0]=1;    int que[MAXN];    int front,rear;    front=rear=0;    dep[end]=0;    que[rear++]=end;    while(front!=rear)    {        int u=que[front++];        if(front==MAXN)front=0;        for(int i=head[u];i!=-1;i=edge[i].next)        {            int v=edge[i].to;            if(dep[v]!=-1)continue;            que[rear++]=v;            if(rear==MAXN)rear=0;            dep[v]=dep[u]+1;            ++gap[dep[v]];        }    }}int SAP(int start,int end){    int res=0;    BFS(start,end);    int cur[MAXN];    int S[MAXN];    int top=0;    memcpy(cur,head,sizeof(head));    int u=start;    int i;    while(dep[start]<n)    {        if(u==end)        {            int temp=INF;            int inser;            for(i=0;i<top;i++)               if(temp>edge[S[i]].cap)               {                   temp=edge[S[i]].cap;                   inser=i;               }            for(i=0;i<top;i++)            {                edge[S[i]].cap-=temp;                edge[S[i]^1].cap+=temp;            }            res+=temp;            top=inser;            u=edge[S[top]].from;        }        if(u!=end&&gap[dep[u]-1]==0)//出现断层，无增广路          break;        for(i=cur[u];i!=-1;i=edge[i].next)           if(edge[i].cap!=0&&dep[u]==dep[edge[i].to]+1)             break;        if(i!=-1)        {            cur[u]=i;            S[top++]=i;            u=edge[i].to;        }        else        {            int min=n;            for(i=head[u];i!=-1;i=edge[i].next)            {                if(edge[i].cap==0)continue;                if(min>dep[edge[i].to])                {                    min=dep[edge[i].to];                    cur[u]=i;                }            }            --gap[dep[u]];            dep[u]=min+1;            ++gap[dep[u]];            if(u!=start)u=edge[S[--top]].from;        }    }    return res;}int main(){     int i,j,k;     int m;    while(scanf("%d%d",&m,&n)==2)    {        init();        int t;        int x,y;        while(m--)        {            scanf("%d%d%d",&x,&y,&t);            addedge(x,y,t);        }        int  source=1;        int sink=n;        t=SAP(source,sink);        printf("%d\n",t);    }    return 0;}